Questions & Answers

Question

Answers

$\begin{align}

& \text{A}\text{. nRT} \\

& \text{B}\text{. }\dfrac{RT}{n} \\

& \text{C}\text{. RT}\left( \text{1-}\dfrac{1}{n} \right) \\

& \text{D}\text{. RT(n-1)} \\

\end{align}$

Answer
Verified

PV = nRT, here P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature of the gas.

In the above graph we see that, the pressure is falling from p to $\dfrac{P}{n}$, and as the pressure falls, the volume also decreases to āVā.

We know that,

${{P}_{1}}{{V}_{1}}=nR{{T}_{1}}$ and ${{P}_{2}}{{V}_{2}}=nR{{T}_{2}}$,

We also know that this ${{T}_{1}}$and${{T}_{2}}$, both are the same.

So we can write,

${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$.

Now if we consider${{P}_{1}}{{V}_{1}}=PV$, then

$PV=\dfrac{P}{n}{{V}_{2}}$, as ${{P}_{2}}=\dfrac{P}{n}$ (given).

${{V}_{2}}=nV$.

Now we have to find the net heat,

We know that net heat is= nRT.

Now as off in the first condition:

$nRT=P\Delta V=\dfrac{P}{n}\left( nV-V \right)$,

$\dfrac{P}{n}V\left( n-1 \right)$, now if we replace the PV with nRT, as PV=nRT.

Then we get the result as,

$RT\left( n-1 \right)$,

An isochoric process is a thermodynamic process in which the volume of the gas remains constant in the system, whereas in the case of an isobaric process the pressure of the system always remains constant.